$f(t) = -4t^{3}-2t^{2}+t+3(h(t))$ $h(x) = -3x^{2}-2x$ $ h(f(0)) = {?} $
First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = -4(0^{3})-2(0^{2})+3(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = -3(0^{2})+(-2)(0)$ $h(0) = 0$